GATE Questions and Answers with Explanation

Q:

A Shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net Profit or loss that he makes?

A) 4%	B) 5%
C) 6%	D) 8%

Answer & Explanation Answer: A) 4%

Explanation:

SP of first article = Rs.1000

Profit = 20%

$C P = S P x \frac{100}{100 + P %} = R s . \frac{2500}{3}$

SP of Second Article = Rs.1000

Loss = 20%

$C P = S P x \frac{100}{100 - L %} = R s . 1250$

So, Total SP = Rs.2000; Total CP = Rs.6250/3

As the CP is more than SP, he makes a loss.

Loss = CP - SP = (6250/3) - 2000 = Rs.(250/3)

So, Loss Percent = Lossx100/CP = 4%

Report Error

View Answer Report Error Discuss

Filed Under: Profit and Loss
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

79 33458

Q:

The Value of $[\log (\tan (1^{0})) + \log (\tan (2^{0})) + \dots \dots + \log (\tan (89^{0}))]$ is

A) -1	B) 0
C) 1/2	D) 1

Answer & Explanation Answer: B) 0

Explanation:

$= [\log \tan (1^{0}) + \log \tan (89^{0})] + [\log \tan (2^{0}) + \log \tan (88^{0})] + \dots \dots + [\log \tan (45^{0})]$

$= \log [\tan (1^{0}) \times \tan (89^{0})] + \log [\tan (2^{0}) \times \tan (88^{0})] + \dots \dots + \log (1)$

$[∵ \tan (90 - θ) = c o t θ a n d \tan (45^{0}) = 1]$

= log 1 + log 1 +.....+log 1

= 0.

Report Error

View Answer Report Error Discuss

Filed Under: Logarithms
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO

27 5406

Q:

A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96	B) 19/90
C) 19/96	D) None of these

Answer & Explanation Answer: B) 19/90

Explanation:

$A S S I S T A N T \to A A I N S S S T T$

$S T A T I S T I C S \to A C I I S S S T T T$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{1_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{9_{C_{1}}} \times \frac{2_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing S = $\frac{3_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/10

Probability of choosing T = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/15

Hence, Required probability = $\frac{1}{45} + \frac{1}{45} + \frac{1}{10} + \frac{1}{15} = \frac{19}{90}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

97 21265

Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A) 8/39	B) 15/39
C) 12/13	D) None of these

Answer & Explanation Answer: B) 15/39

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

= $1 - (\frac{16_{C_{1}} \times 14_{C_{1}} \times 12_{C_{1}} \times 10_{C_{1}}}{16_{C_{4}}}) = \frac{15}{39}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

64 13681

Q:

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A) 23/42	B) 19/42
C) 7/32	D) 16/39

Answer & Explanation Answer: B) 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore

P(E1) = 1/2 and P(E2) = 1/2

Now, $P (\frac{A}{E_{1}})$ = Probability of drawing a red ball when the first bag has been selected = 4/7

$P (\frac{A}{E_{2}})$ = Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have

P(red ball) = P(A) = $P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}})$

= $\frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{2}{6} = \frac{19}{42}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO

64 26717

Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7	B) 1/5
C) 2/7	D) 2/35

Answer & Explanation Answer: C) 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)]

= $P [(E \cap F) \cup (F \cap E)]$

= $P (E) P (F) + P (F) P (E)$

= $\frac{1}{7} \times \frac{4}{5} + \frac{1}{5} \times \frac{6}{7} = \frac{2}{7}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

17 5468

Q:

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

A) 1/4	B) 1/2
C) 3/4	D) 7/12

Answer & Explanation Answer: C) 3/4

Explanation:

Let A, B, C be the respective events of solving the problem and $A, B, C$ be the respective events of not solving the problem. Then A, B, C are independent event

$∴ A, B, C$ are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$P (A) = \frac{1}{2}, P (B) = \frac{2}{3}, P (C) = \frac{3}{4}$

$∴$ P( none solves the problem) = P(not A) and (not B) and (not C)

= $P (A \cap B \cap C)$

= $P (A) P (B) P (C)$ $[∵ A, B, C a r e I n d e p e n d e n t]$

= $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}$

= $\frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1 - \frac{1}{4}$ = 3/4

Report Error

View Answer Report Error Discuss

Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

909 196867

Q:

A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

A) 123	B) 113
C) 246	D) 945

Answer & Explanation Answer: C) 246

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.

(i) 1 lady out of 4 and 4 gentlemen out of 6

(ii) 2 ladies out of 4 and 3 gentlemen out of 6

(iii) 3 ladies out of 4 and 2 gentlemen out of 6

(iv) 4 ladies out of 4 and 1 gentlemen out of 6

In case I the number of ways = $C_{1}^{4} \times C_{4}^{6}$ = 4 x 15 = 60

In case II the number of ways = $C_{2}^{4} \times C_{3}^{6}$ = 6 x 20 = 120

In case III the number of ways = $C_{3}^{4} \times C_{2}^{6}$ = 4 x 15 = 60

In case IV the number of ways = $C_{4}^{4} \times C_{1}^{6}$ = 1 x 6 = 6

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

10 11384

GATE Questions

Quick Links